jasperE3 wrote: If $a,b$ are nonnegative real numbers, prove the inequality $$\frac{a+\sqrt[3]{a^2b}+\sqrt[3]{ab^2}+b}4\le\frac{\sqrt{a+\sqrt{ab}+b}}3.$$ Let $a,b>0$. Prove that $$(\frac{\sqrt{a}+\sqrt{b}}{2})^2\leq \frac{a+\sqrt[3]{a^2b}+\sqrt[3]{ab^2}+b}{4}\leq\frac{a+\sqrt{ab}+b}{3}\leq\sqrt{(\frac{\sqrt[3]{a^2}+\sqrt[3]{b^2}}{2}^3)}$$